SOME PROBLEMS WITH SOLUTIONS

Transcription

1 SOME PROBLEMS WITH SOLUTIONS FMSN5/MASM4 VALUATION OF DERIVATIVE ASSETS MAGNUS WIKTORSSON Augus 013 Faculy of Engineering Cenre for Mahemaical Sciences Mahemaical Saisics CENTRUM SCIENTIARUM MATHEMATICARUM

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3 Chaper 1 Problems 1.1 Maringales Assume ha he process {S } 0 follows he sandard Black & Scholes model and haγ R. Findγ 1 such ha {S γ e r } 0 will be a Q-maringale Show ha he process X = e / cosw, where W is a sandard Brownian moion, is a maringale for Saic replicaion of derivaives 1..1 We have a derivaive wih mauriy T and pay-off ΦS T = maxk K S T, 0. K pay off funcion Φx 0 0 K K x Find a saic hedge for his derivaive using he asse S and European opions on S. 1.. Assume ha X is a derivaive wih mauriy T having he following pay- off funcion: K S T K K S T K S T K 0 S T K Express he price for 0 < < T of X using European pu and call opions, he sock S and he bank-accoun B. 1

4 1.3 PDE:s and Feynman-Kac formula Solve he PDE f, x + μ σ f, x + σ f, x x x = 0 f T, x = e x for [0, T, x R using he Feynman-Kac represenaion, whereμandσare real-valued consans. 1.4 Pricing of derivaives and hedging Consider a derivaive wih pay-offφs T = maxk S T, 0 in he Black-Scholes marke. a Find he price-formula for he derivaive for 0 T. b Calculae he replicaing porfolio dela-hedge for he derivaive, ha is find a self-financing porfolio consising of he sock and he bank accoun ha hedges he derivaive. 1.5 Change of measure and compleeness A simple exension o he binomial model is he rinomial model TM where we also allow he sock price o remain unchanged wih some probabiliy. More precisely we have ha PS k+1 = us k = p u, PS k+1 = S k = p 1, PS k+1 = ds k = p d, k = 0, 1,, N 1 in he N -period case where u > 1 and d < 1 and a bank accoun B wih B k = expkr. Even hough i can provide a beer approximaion of he Black-Scholes model for a finie number of periods, TM is seldom used. The main reason for his i ha TM gives an incomplee marke. We are here going o sudy TM in he one period seing wih u = 6/5, d = 3/5, S 0 = 1 and wih zero ineres rae. a Show ha TM is incomplee by showing ha he risk-neural probabiliies q u, q 1 and q d are no unique. b If we add anoher asse o he marke i is possible o make he marke complee. Suppose ha we add he asse X where X is he derivaive wih pay-off S 1 a ime 1 and furher suppose ha he price of X a ime zero is equal o 6/5. If he price of X is given by he marke we can view his as a calibraion of our model o fi he marke price. This will hen allow us o price oher derivaives on he asse S. Show ha he marke is complee by finding he unique risk-neural probabiliies. c Use he probabiliies from b o price he derivaive Y, wih mauriy T = 1 and pay-off maxs 1 9/10, 0. Moreover find he replicaing porfolio for his derivaive using he asses S, X and B.

5 d Noe ha we canno have an arbirary price of X a ime zero. Show ha all prices ouside he inerval [1, 7/5 will give arbirage in he sense ha here are no risk-neural probabiliies compaible wih hose prices Assume ha we have a marke consising of one risky asse S 1 and one bank accoun S 0 wih hep-dynamics: ds 1 = μ 1 S 1 d + S 1 σ11 dw 1 +σ 1 dw ds 0 = rs 0 d, where W 1 and W are wo independen sandard BM:s. Using he mea-heorem on his model we ge ha i is free of arbirage bu incomplee, i.e. he maringale measure exiss bu is no unique. a However, show ha all simple claims X wih mauriy T of he formφs 1 T will have a price-formula ha does no depend on he choice of maringale measure. b Show ha he claim Y = 1 W T >K, ha is a conrac which pays one uni of currency a ime T if W T > K, will have a price-formula ha do depend on he choice of maringale measure. c Show ha if we add he asse S wih P dynamics: ds =μ S d +σ S dw o he marke, ha he marke will be free of arbirage and complee by finding he unique Girsanov kernel g 1, g. d Price he derivaive in b using his unique maringale measure. 1.6 Change of numeraires Assume he following -dim Black-Scholes model under Q for he wo socks S 1 and S, ds 1 ds = rs 1 d + S 1 σ 11 dw 1 +σ 1 dw = rs d + S σ 1 dw 1 +σ dw where W 1 and W are wo independen sandard Q Brownian moions and where r,σ 11,σ 1,σ 1 andσ are posiive consans. Price he derivaive wih mauriy T and pay-off: ΦS 1 T, S T = maxs T S 1 T, 0, for 0 < < T. Hin: he key is o find he righ numeraires In a realisic siuaion he shor ineres rae r is no a deerminisic consan. Wha one wans o do is o use observed prices of Zero Coupon bonds ZCB as a discouning facor when pricing derivaives. The way o accomplish his is o express he dynamics of he underlying sock S under he forward measure Q T, i.e. he maringale measure which has he ZCB as numeraire. Under he measure Q T we 3

6 have ha he discouned sock process Z = S/p, T should be a maringale, where p, T is he price a ime of a ZCB wih mauriy T. Noe ha We assume he following model for Z under Q T : dz = Zv, T dw QT, 0 T, where W QT is a sandard -dim Q T Brownian moion and where v, T is a posiive deerminisic funcion a 1 row vecor. Noe ha by definiion we have ZT = ST since pt, T = 1. a Now assume ha under he usual maringale measure Q, we have he following model for he ZCB and he sock, dp, T = rp, T d + P, T T γdw 1, 0 T, ds = rsd +σsdw, where W 1 and W are wo independen sandard Q Brownian moions and whereσandγare posiive consans. Calculae he volailiy funcion v, T implied by his model. Recall ha he volailiies do no change when we change he measure. b Price a sandard European call opion wih srike K and mauriy T for his model, for such ha 0 < < T. You should express he price in a Black- Scholes ype of formula. Remember o check ha your price simplifies o he sandard Black-Scholes formula if he ineres rae is consan equal o r and v, T σ. 1.7 Volailiy VIX: On Chicago Board Opions Exchange CBOE here is an index called VIX, which measures expeced volailiy on he SP500 index. The VIX is calculaed using marke prices of European pu and call opions on he SP500 index. Le ΠE P, +τ, S, K andπ E C, +τ, S, K be prices of a European pu and a European call opions respecively given a ime wih mauriy + τ wih curren sock price S and wih srike K. Le fu rher F +τ be he forward level a ime of he sock price a ime +τ, i.e. F +τ = E Q [S +τ F. Having access o a wide range of opion prices over differen srikes we can ge an almos model-free esimae of he expeced forward volailiy. VIX,τ = NP 1 e rτ ΠE P, +τ, S, K i K i+1 K i τ Ki IK i < F +τ + τ i=1 NC 1 j=1 e rτ ΠE C, +τ, S, K j K j+1 K j Kj IK j > F +τ where NP and NC are he number of available pu and call opions wih mauriy +τ respecively 1. The empirically calculaed VIX index squared VIX,τ is an 1 This is a slighly simplified version of he real index. 1/, 4

7 approximaion of an inegral VIX,τ F +τ e rτ ΠE P, +τ, S, K dk τ 0 K + e rτ ΠE C, +τ, S, K dk F +τ K [ = F +τ τ EQ K S +τ + dk 0 K + S +τ K + dk F +τ K F = S +τ [ τ EQ ln F +τ F. Now assume ha we have a marke wih he following Q-dynamics: db ds = rbd, = rsd + σsdw, where r is he risk free rae,σadeerminisic funcion and W is a sandard Q- Brownian moion. This is he model you shall use in he res of he problem. a Calculae F +τ = E Q [S +τ F. b Evaluae he expecaion c Show ha [ F +τ E Q 0 τ EQ [ ln K S +τ + dk K + S +τ F +τ F. F +τ S +τ K + dk K 1.7. VIX con. Assume ha S follows he Heson model i.e. we have a marke wih he following Q-dynamics: db ds dv = rbd, F = rsd + V SρdW ρ dw, = κθ V d +σ v V dw1, where r is he risk free rae,σ v,κ,θ are posiive deerminisic consans, 1 ρ 1 and W 1 and W are independen sandard Q-Brownian moions. Evaluae he expecaion [ τ EQ ln where F +τ = E Q [S +τ F. S +τ F +τ F, 1.8 Maringale models for he shor rae Calculae he price of a Zero Coupon bond wih mauriy T a ime where 0 < < T in he following Ho-Lee model for he shor rae, { drs =Θs ds+σdw s, for s r = r whereθs = ce s +σ s and whereσ, r and c are posiive consans. [ S +τ = E Q ln F +τ F. 5

8 1.9 The HJM framework Consider he following HJM model for he forward rae f, T, T 0; df, T = σ T d +σdw Q, 0 T f 0, T = f 0, T, T 0 where W is a sandard BM and f 0, T is he observed forward rae on h e marke. a Find he ZCB price p, T for his model. b Calculae heq S -dynamics, ha is he dynamics under he numeraire measure where p, S is used as a numeraire, for he process X = 1+S T L [T, S for 0 T, where L [T, S = p, T p, S/S T p, S. c Use he dynamics obained in b o price he Caple for 0 T, ha is he derivaive wih mauriy S and pay-off where K is a posiive consan. S T maxl T [T, S K,0, 6

9 Chaper Soluions Sol: 1.1 Maringales Sol: Using ha S has Q-dynamics ds = rs d +σs dw and applying Iô s formula o he process X = S γ e r we ge ha dx = rx +γrx + 1 σ γγ 1X d +γσx dw = X rγ 1+ 1 σ γγ 1d +γσx dw = X γ 1r + 1 σ γd +γσx dw. For X o be a maringale we need ha he drif is zero which gives haγshould solve he equaion γ 1r + 1 σ γ = 0. We see righ away ha i has he soluionsγ = 1 andγ = r σ and since we waned a soluion differen fromγ = 1 we see ha he requesed soluion isγ = r σ. We should also check ha X saisfies he required momen condiions. Plugging in γ = r σ we ge he following dynamics for X: which has he soluion Now we have assuming X 0 = S γ 0 > 0 dx = r σ X dw, X = X 0 exp r σ r σ W. E[ X = E[X = E[X 0 exp r σ r σ W = X 0 exp r σ +r σ = X 0 <, which shows ha X is a maringale under Q. 7

10 Sol: 1.1. Le X = f, W = e / cosw. We now calculae dx wih Iôs formula: { f dx = df, W = + 1 } f x d + 1 f x dw { 1 = e/ cosw 1 } e/ cosw d e / sinw dw = e / sinw dw This process has no drif erm moreover we have ha he diffusion par has finie second momen using he Iô isomery, i.e. E[ 0 e s/ sinw s dw s = This gives ha X is a Maringale. 0 E[e s sinw s ds Alernaive soluion: We can direcly calculaee[x F s for s < as 0 E[e s ds = e 1 <. E[X F s = E[e / cosw F s = E[e s/ cosw W s + W s e s/ F s = E[e s/ cosw W s cosw s e s/ sinw W s sinw s e s/ F s = e s/ cosw s e s/ e s / +sinw s e s/ 0 = e s/ cosw s = X s. Finally we need o esablish ha E[ X < which is easily done since X < e / <. Sol: 1. Saic replicaion of derivaives Sol: 1..1 The easies way o find a hedge for he pay-off is o draw a picure. K 0 S maxs K,0 maxs K,0 0 K K S The solid line correspond o he original pay-off and he dashed lines correspond o he sock, and one long European call opion wih srike K and wo shor European 8

11 call opions wih srike K respecively op o boom. So we can replicae he payoff using one sock, and one long European call opion wih srike K and wo shor European call opion wih srike K. To do hings properly we should check ha S maxs K,0+maxS K,0 = maxk S K, 0 for all S 0. We sar wih he lef hand side S 0 S K S maxs K,0+maxS K,0 = K S K S K 0 S K and hen move on o he righ hand side S 0 S K maxk S K, 0 = K S K S K 0 S K which shows he equivalence beween he wo pay-offs. Sol: 1.. LeΠ be he price of he derivaive X for T, moreover le P K and C K be he price a ime of a pu opion and call opion respecively boh wih mauriy T and srike price K. The priceπ is given by any of he following equivalen porfolios here are more possible equivalen porfolios. Π = P K P K = K B/BT C K +C K = S+P K C K +C K Looking a he payoffs a ime T show ha he porfolios have he same payoff as he conrac X, i.e. K ST K K ST K ST K 0 ST K Sol: 1.3 PDE:s and Feynman-Kac formula Sol: According o Feynman-Kac s represenaion heorem he PDE is solved by where X has he dynamics f, x = E[e X T X = x, dx s = μ σ ds+σdw s, s T X = x. 9

12 I is sraighforward o see a leas i should be ha X is is a BM wih drifμ σ and sandard deviaionσsaring a x a ime, i.e. X T = x + μ σ T +σw T W. Looking a expx T we see ha i has exacly he same disribuion as he geomeric BM in he sandard Black-Scholes model. We hus ge ha f, x = E[e X T X = x = E [e x e μ σ T +σw T W = e x+μ σ T +σ T = e x+μt. We should also check he obained soluion fulfills he PDE and he boundary condiion. We sar wih he las ask f T, x = e x+0μ = e x as prescribed. Finally we ge ha f, x + μ σ f, x + σ f, x x x which verifies ha he obained soluion is correc. = μf, x+μ σ f, x+σ f, x = 0, Sol: 1.4 Pricing of derivaives and hedging Sol: a Under Q we have ha S T = S e r σ T +σw T W. According o he risk-neural-valuaion-formula we have ha he price of he derivaive,π = F, S, is given by F, S = e rt E Q[ Markov ΦS T F = e rt E Q[ ΦS T S = e rt E Q[ maxk S T, 0 S = e rt E Q[ K S T 1 ST K S [ = e rt E Q K S e r σ T +σw T W 1 e r σ T +σw T W S K/S [ = e rt E Q K S e r σ T +σ T Z S 1Z log K/S r σ, where Z N0, 1. Expressing he expecaion as an inegral we ge T σ T F, S = e rt K = e rt K = e rt K = e rt K d1 d1 d1 d1 e z d1 dz S e r σ T π e z d1 dz S e r σ T π e z d1 dz S e r+σ T π e z π dz S e r+σ T e σ T z z π dz e 1 4σ T z+z +4σ T 4σ T dz π e 1 z σ T π dz d1 σ T = e rt K Nd 1 S e r+σ T Nd 1 σ T, e 1 z π dz 10

13 where N is he sandard normal disribuion funcion and where d 1 = log K/S r σ T σ. T b Le h = h b, h s be a self-financing porfolio wih value process V = h b B + h s S. By he self-financing condiion we ge ha he dynamics of V is given by dv = h b db + h s ds. We now wan o choose h b and h s such V and he derivaive wih price process Π has he same dynamics. The dela-hedge gives ha we should choose h s as h s = S F, S and since we should have V = F, S we ge ha h b = F, S h s S B Using he F obained in a we ge ha = F, S S S F, S. B h s = S e r+σ T Nd 1 σ T h b = e rt K Nd 1 +S e r+σ T Nd 1 σ T e r. Alernaive derivaion: If you have forgoen he dela-hedge you should look a following o refresh your memory. Applying Iô s formula oπ = F, S we ge ha dπ = F, S + σ S S F, S d + F, S ds S rb = F, S = F, S + σ S F, S + σ S S S So if we choose h b = F, S + σ S h s = F, S S d + F, S ds rb S 1 F, S db + F, S ds. rb S S 1 F, S rb we ge ha V andπ will have he same dynamics. We can simplify his furher by using ha V =Π = F, S which also follows from ha F saisfies he Black-Scholes PDE so ha we ge ha h b = F, S h s S B = F, S S S F, S B 11

14 Sol: 1.5 Change of measure and compleeness Sol: The mahemaics in his soluion is only simple linear algebra bu he financial implicaions are sill noeworhy. a The condiion for he risk neural probabiliies are ha S 0 = E Q [S 1 S 0 = q d ds 0 + q 1 S 0 + q u us 0 and ha he risk neural probabiliies sum o one. We mus also have ha he probaliies are posiive and smaller han one. We hus obain he following linear sysem of equaions for he risk neural probabiliies: [ /5 1 6/5 q d q 1 q u = [ 1 1 So we have wo equaions and hree unknowns so here exiss muliple infiniely many soluions. In principle we are done here since now we have ha he risk neural probabilies are no unique. We should check ha a leas wo soluions saisfy ha hey are non-negaive and smaller han one. We however need he full soluion in d so we solve i now. We sar by finding all soluions and hen we make resricions so ha he probaliies are posiive and smaller han one. Three unknowns and wo equaions gives us a one parameric family of soluions. We pu q d =θ and aim o express he oher probaliies usingθ. Plugging his ino he our equions above we obain [ /5 [ q1 q u [ = 1 θ 1 3/5θ This has he soluion [q 1, q u = [1 3θ, θ. We here imediaely see ha 0 θ 1/3 for his o be probabiliies. For financial reasons we do no wan he casesθ = 0 implies S = B wih prob 1 andθ = 1/3 which gives us back he bionomial model wih prob 1. Taking his ino consideraion gives us he soluions [q d, q 1, q u = [θ, 1 3θ, θ, 0 <θ<1/3. We can hus by changingθcalibrae o differen marke siuaions. b Since we all ready have he full soluion all we need o do here is o findθso ha he price of he derivaive mach, i.e. we calibrae our model o fi he marke price. We have ha E Q [S 1 S0 = 1 = 6/5 which gives ha.. θ3/ θ1 + θ6/5 = 6/5, 1+θ9 75+7/5 = 6/5, 1+6θ/5 = 6/5 6θ = 1 θ = 1/6. So we have a unique risk nerual measure wih [q d, q 1, q u = [1/6, 1/, 1/3. c We now use he probabiliies from b o price he derivae E Q [maxs 1 9/10, 0 S 0 = 1 = 0 1/6+1 9/10 1/+6/5 9/10 1/3 = 1/0+1/10 = 3/0. 1

15 To find he hedge porfolio we mus solve sysem of linear equaions jus as in he binomial model bu now we have hree asses and hree equaions. So we should find h X, h S and h B which are he porfolio weighs for X = S, S and B respecively such ha h X S 1 + h S S 1 + h B B 1 = maxs 1 9/10, 0, for all possible oucomes of S 1. This gives rise o following sysem of equaions 9/5 3/5 1 h X h S = 1/10 36/5 6/5 1 h B 3/10 We easily solve his by Gauss eliminaion obaining h X 5/4 h S h B = 7/4 3/5. d From a and b we ge ha E Q [S 1 S0 = 1 = 1+6θ/5, 0 <θ<1/3. This is an increasing funcion ofθhus we ge he upper and lower bounds for he price by plugging in he allowed upper and lower bounds forθ. This gives ha 1 < E Q [S 1 S 0 < 7/5, which was o be shown. So i is no possible o calibrae he model o prices ouside his inerval, since ha would imply negaive probaliies and or probabiliies larger han one. Sol: 1.5. a Regardless of he choice of maringale measure Q we have ha S 1 will have heq-dynamics ds 1 = rs 1 d + S 1 σ11 dw 1,Q +σ 1 dw,q, where W 1,Q and W,Q are sandard BM:s under Q. We herefore ge ha S T = S e r σ 11 +σ 1 which has he same disribuion as T +σ 11W 1,Q T W 1,Q +σ 1W,Q T W,Q, S e r σ 11 +σ 1 T + σ σ T Z, where, Z N0, 1. Thus he price a ime,π, of a simple claim wih mauriy T and pay-off ΦS T is given by Π = e rt Φ S e r σ 11 +σ 1 T + σ 11 +σ 1 regardless of he choice of Q, which was o be shown. T z e z dz π 13

16 b The dynamics of W under Q is given by dw = dw,q g d where g 1, g are any funcions saisfying he equaion μ 1 σ 11 g 1 σ 1 g = r, for, 0 T and he Novikov condiion [ E P e T g 1 s +g s 0 ds <. Two possible choices are e.g. g 1, g = μ 1 r/σ 11, 0 and g 1, g = 0, μ 1 r/σ 1. These wo choices will give differen prices o he derivaive 1 W T >K, more precisely W e rt K N T and e rt N W K μ1 r σ 1 T T respecively. c We ge ha g 1, g should solve he following sysem of linear equaions σ 11 g 1 σ 1 g = r μ 1 σ g = r μ which has he unique soluion provided haσ 11 σ 0 g = μ r, g 1 = μ μ 1 r σ r 1 σ, σ σ 11 and since he Girsanov kernel is unique we ge ha he marke is free of arbirage and complee. d Using he resul from c we ge ha he price of he derivaive is given as W e rt K μ r σ N T. T Remark: Noe ha usually we say ha a derivaive or coningen claim should only be a funcion of he rajecory of he underlying asse S 1 up o mauriy. The derivaive in a is such a conrac and here i does no maer ha we have added an exra Browninan moion. However he conrac in b is no a funcion of S 1 only, here we need more informaion o know he value a mauriy. This ambiguiy regarding compleeness is he reason why i is called a mea-heorem see Åberg p. 111 and Björk p. 1 and no a heorem. The moral of his is perhaps ha we should use a model descripion ha only uses as many driving Brownian moions as we acually need o represen he model. 14

17 Sol: 1.6 Change of numeraires Sol: We wan o price he derivaive X wih mauriy T and pay-off: ΦS 1 T, S T = maxs T S 1 T, 0. According o he risk-neural valuaion formula he price a ime,π X, is given by Π X = e rt E Q [maxs T S 1 T, 0 F. There are now some differen approaches o calculae his expecaion using change of numeraire echniques. The perhaps easies approach is o use S 1 as a numeraire his leads o [ [ Π X = S 1 E QS 1 1 S 1 T maxs T S 1 T, 0 F = S 1 E QS 1 max S T S 1 T 1, 0 F, whereq S1 is he numeraire measure for S 1. This can now be seen as a European call opion on he raio S /S 1 wih srike 1. Moreover since S /S 1 is a raio beween a raded asse and a numeraire i is auomaically a maringale under Q S1. Using ha he volailiies does no change when we change measure we can by calculae he volailiies under Q as S S S 1 σ 11 dw 1 +σ 1 dw + S σ 1 dw 1 +σ dw S 1 S 1 S S 1 = S S 1 σ 1 σ 11 dw 1 +σ σ 1 dw This gives ha he dynamics for S /S 1 underq S1 for s is given by d S s S 1 s S S 1 = S s σ 1 σ 11 dw QS 1 1 s+σ σ 1 dw QS 1 s, S 1 s = s s 1 where W QS 1 1 s and W QS 1 s are independen sandard Q S1 Brownian moions. We can now solve his SDE o obain S T S 1 T = s s 1 exp 1 σ1 σ 11 + σ σ 1 T +σ 1 σ 11 W QS 1 1 T W QS σ σ 1 W QS 1 T W QS 1 This has he same disribuion as s s 1 exp 1 σ T +σ T G where G is a sandard Gaussian random variable and where σ = σ 1 σ 11 + σ σ 1., 15

18 We can herefore calculae he priceπ X as s Π X = s 1 max s 1 exp s = s 1 d s 1 exp s 1 = s 1 exp d s 1 π s 1 = s 1 exp s 1 π where = s d d 1 π exp 1 σ T +σ T g 1 σ T +σ T g exp g / 1, 0 dg π exp g / 1 dg π 1 σ T +σ T g g / exp g / dg π 1 g σ T exp g / π dg 1 g σ T exp g / dg s 1 dg d π = s 1 N d σ T s 1 1 N d = s N d +σ T s 1 N d, d = lns /s 1 σ T / σ, T and where N is he disribuion funcion of he sandard Gaussian disribuion. Alernaive soluion: We can also use boh S 1 and S as numeraires which gives ha Π X = S E QS [ IS T > S 1 T F S1 E QS 1 [ IS T > S 1 T F. This can now be rewrien as [ Π X = S E QS S1 T I S T < 1 [ F S 1 E QS 1 S T I S 1 T > 1 F. Now S 1 /S is a Q S -maringale and S /S 1 is a Q S1 -maringale. By using he same ype of argumen and calculaions as in he firs soluion we obain ha S 1 T /S T under Q S has he same disribuion as s 1 s exp s s 1 exp 1 σ T +σ T G and ha S T /S 1 T under Q S1 has he same disribuion as 1 σ T +σ T G whereσand G are as defined in he firs soluion. Sraighforward calculaion now give ha Π X = s N d +σ T s 1 N d, where d are defined as in he firs soluion. Alernaive soluion : We can also calculae he dynamics for S 1 and S under boh Q S1 and Q S. For his we need o find wo wo-dimensional Girsanov kernels, where we also need o look a dynamics of he bank-accoun o ge he righ Girsanov kernels. This is,, 16

19 however as seen from he calculaions above an unnecessary deour. For he sake of compleeness we supply he appropriae Girsanov kernels: g S1 1 = σ 11, g S1 = σ 1 g S 1 = σ 1, g S = σ. By he same ype of calculaions as in he firs wo soluions we finally arrive a he same answer. Sol: 1.6. a By using ha Z = S/P, T, for 0 T, is a Q T maringale and ha volailiies do noe change when we change measure, we see ha we can calculae he volailiy funcion v, T using he diffusion par of he Q-dynamics for S/P, T. This gives ha Zv, T [ dw1 dw S = P, T γt dw 1 + P P S = ZγT dw 1 +ZσdW. = Z [ γt σ [ dw 1 dw S SσdW P This gives ha v, T = [ γt σ. b Using ha ST = ZT we can view he conrac as wrien on he process Z insead of S. So he price of he European call opion can hus be calculaed as Π = p, T E QT [ maxzt K,0 F. To calculae his we mus find he disribuion of ZT under Q T. Solving he SDE for Z under Q T gives ha ZT = Z exp 1 T T vs, T ds+ vs, T dw QT s. This now has he same disribuion as Z exp 1 Σ, T T +Σ, T T G, where Σ, T = T γ T s +σ ds T γ T = 3 /3+σ T, T and where G is sandard Gaussian random variable. By exacly he same calculaions as in he derivaion of he sandard Black-Scholes formula bu wih σ replaced byσ, T and e rt replace by p, T we obain Π = p, T ZN d +Σ, T T p, T KN d = SN d +Σ, T T p, T KN d, 17

20 where d = lnz/k Σ, T T / Σ, T T = lns/k lnp, T Σ, T T / Σ, T, T and where N is he disribuion funcion of he sandard Gaussian disribuion. To check ha we ge back he original formula if r r and v, T σ, we noice ha his imply haγ = 0 and p, T = exp rt. This gives haσ, T σ and lnp, T = rt. Now plugging his ino our price gives Π = SN d +σ T e rt KN d, where d = lns/k +rt σ T / σ, T which we recognize as he ordinary Black-Scholes formula. Sol: 1.7 Volailiy Sol: a Using ha we ge ha e r+τ E Q [S +τ F = S F,+τ = E Q [S +τ F = e rτ S. b By applying Io s formula o lnsu we ge ha d lnsu = r σ u/du+σudw u. We solve his by direc inegraion o obain lns +τ = lns+rτ +τ σ u/du+ Using ha he las erm is a maringale we obain E Q [lns +τ F = lns+rτ This ogeher wih he resul in a hen gives ha [ E Q,+τ lns +τ/f F τ = τ = 1 τ +τ +τ σudw u. σ u/du. lns+rτ 1 +τ σ udu. +τ So here we see ha he VIX-index squared really is an approximaion of he average squared fuure volailiy. σ udu lns rτ 18

21 c To ease he noanal complexiy we pu S = S +τ and F = F +τ. We sar by direc calculaion of he inegrals F 0 K S + dk K + F S K + dk K = = = F S F K S dk IS < F+ K 1 S K S dkis < F+ K [ lnk + S F K IS < F+ S S F S F S K dk IF < S K S K 1 K dkif < S [ S F K lnk IF < S S = lnf lns+ S F 1IS < F+IF < s Taking condiional expecaion and using he definiion of F +τ E Q [ lns+τ/f +τ + which shows ha S +τ F +τ [ F +τ E Q K S +τ + dk K + 0 = lnf lns+ S F 1 = lns/f+ S F 1. 1 F = E Q [ lns+τ/f +τ F +τ S +τ K + dk K we obain F F + F +τ F +τ 1 = E Q [ lns+τ/f +τ F, [ S +τ = E Q ln F +τ F. Sol: 1.7. By applying Io s formula o lnsu we ge ha d lnsu = r V u/du+ V udw 1 u+ 1 ρ dw u. We solve his by direc inegraion o obain lns+τ = lns+rτ +τ V u/du+ Using ha he las erm is a maringale we obain +τ E Q [lns +τ F = lns+rτ E Q [ +τ This ogeher wih he ha F +τ [ E Q,+τ lns +τ/f F τ = e rτ S hen gives ha V udw1 u+ 1 ρ dw u. V u/du F. = [ 1 lns+rτ E Q τ = 1 [ +τ τ EQ V udu F. +τ Now o calculae his we need o use he dynamics of V. Using direc inregraion we see ha we can represen V u, u > as V u = V + u κθ V sdss+ 19 u V sσv dw 1 s. V udu F lns rτ

22 Taking condiional expecion on boh sides using ha he las erm is a maringale we obain E Q[ [ u [ u V u F = E Q V + κθ V sds F = V +κθu κe Q Now pu mu = E Q[ V u F we hen ge mu = V + u κθ msds. Taking derivaives w.r. u of boh sides we obain he ODE ṁu =κθ mu m = V. Making he change of varables mu =θ mu we obain a sandard lineear ODE We solve his sraigh away yielding and hen we ge ha mu = κ mu m =θ V. mu = θ V e κu mu =θ θ V e κu. V sds F. We can now finally calculae [ 1 +τ τ EQ V udu F = 1 τ = 1 τ = θ +τ +τ mudu θ θ V e κu du θ V 1 e κτ κτ = V 1 e κτ κτ +θ e κτ 1+κτ κτ Noe ha his expression ends o V asτends o zero and i ends oθasτends o infiniy. This makes sense since V is he curren squared volailiy andθis he long erm mean of squared volailiy. Sol: 1.8 Maringale models for he shor rae Sol: The price of he ZCB, p, T, is given by p, T = E [e T rsds. 0

23 The shor rae rs for s > is given by rs = r+ s Θudu+ s σdw u = r ce s e + σ s +σ The inegral of he shor rae rs beween and T is hen given by T T rsds = r ce s e + σ s +σ = rt ce T e ce T σ 6 T σ T σ The expecaion of he sochasic erm is [ T s E dw uds = 0, and he variance is [ V T s dw uds = E T = E = T T s T u dw uds s s dw u. dw u ds T s dw uds. T dsdw u = E T udw u T T u T u3 T 3 du = [ =. 3 3 We see ha he sochasic erm has a normal disribuion, and herefore T s T 3 σ dw uds N 0,σ. 3 Moreover we have ha This gives ha p, T is given by E [e σ T s dw uds = e σ 6 T 3. p, T = exp { rt ce T e ce T σ 6 T σ } T +σ T 3 6. Sol: 1.9 The HJM framework Sol: a Using ha f, u = f 0, u + σ u sds+ σdws Q 0 0 = f 0, u σ u u +σw Q 1

24 and ha we ge ha p, T = e T = e T = e T p, T = e T f 0,udu+ σ f 0,udu+ σ f,udu T u u du σ T W Q du T 3 T T σw Q f 0,udu σ T T T σwq. Noe ha p 0, T /p 0, = exp T f 0, udu and ha r = f, = f 0, + σ /+σw Q, so we can furher simplify he expression as p, T = p 0, T T p 0, exp f 0, σ T T r, which is he Ho-Lee model s ZCB-price when calibraed o he iniial forward curve. This form is however, no so well suied for furher calculaions. b Using he definiion of X and L [T, S we see ha p, T p, S X = 1+S T = p, T S T p, S p, S. If we now use he resul from a we ge ha X = e T = e S = e S f 0,udu σ T T T σwq T f 0,udu+ σ SS T T +S T σwq T f 0,udu σ S T S+T +S T σwq + S f 0,udu σ SS +S σwq To find he dynamics under Q S we noe ha X is he raio of he raded asse p, T and he numeraire p, S herefore we mus have ha X is a maringale under Q S. Therefore we only need o calculae he diffusion par of he dynamics since we know ha he drif par mus be zero, doing his we obain ha X has he following dynamics under Q S where W QS dx =σs T X dw QS, is a sandard BM under Q S. This gives ha X T = X e σ S T T +S T σw QS T WQS c We sar by observing ha we can express he pay-off in erms of X insead of L [T, S giving ha XT 1 S T maxl T [T, S K,0 = S T max S T K,0 = maxx T 1+S T K, 0. This can now be seen as a sandard European call opion on X T wih srike level 1+S T K, excep ha we will no ge paid a unil ime S as opposed

25 o ime T in he sandard case. Using he RNVF for he numeraire measure Q S we ge ha he price of he Caple a ime Π for 0 T is given by Π = p, SE QS [ maxx T 1+S T K, 0 F. Using he resul from b we ge ha Π = p, SX Nd 1 p, S1+S T K Nd = p, T Nd 1 p, S1+S T K Nd, where d 1 = log X /1+S T K +S T σ T //S T σ T and d = d 1 S T σ T. 3

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